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Tales by Dots and Lines Exercise 5.1 Class 8 Maths – Chapter 5 NCERT Solutions

NCERT Class 8 Maths Chapter 5 Tales by Dots and Lines Exercise 5.1 detailed solutions prepared by the ALLEN expert team are provided below. These solutions offer clear explanations of the questions and can be useful for exam preparation and revision.The first exercise of Chapter 5 mainly focuses on understanding mean and median through data sets and dot plots. It also explores how changes in data values affect the mean and median, including finding missing values when the mean is given.

1.0Download Class 8 Maths Chapter 5 Ex 5.1 NCERT Solutions PDF

Download the Class 8 Maths Chapter 5 Exercise 5.1 NCERT Solutions PDF for clear solutions and quick revision.

Chapter 5 : Tales by Dots and Lines Exercise 5.1

2.0Detailed NCERT Solutions for Tales by Dots and Lines Exercise 5.1 Class 8 Maths (Ganita Prakash Term 2)

1. Find the mean of the following data and share your observations: (i) The first 50 natural numbers. (ii) The first 50 odd numbers. (iii) The first 50 multiples of 4.

Sol. (i) The mean of the first 50 natural numbers. The first 50 natural numbers 1, 2, 3, ......, 50 The sum of $n$ natural number is $\frac{n ( n + 1 )}{2}$ So, for $n = 50$ The sum $= \frac{50 ( 50 + 1 )}{2}$ $= \frac{50 \times 51}{2} = 1275$ Mean $= \frac{sum of the first 50 natural numbers}{Total number of natural numbers}$ $= \frac{1275}{50} = 25.5$ Observations: The mean of the first n natural numbers is always $(\frac{n + 1}{2})$

For $n = 50, \frac{51}{2} = 25.5$ (ii) Mean of the first 50 odd number.

The first 50 odd numbers 1, 3, 5, 7, 9, ......, 99 Sum of the first n odd numbers $= n^{2}$ So for $n (= 50)$ the sum $= (50)^{2} = 2500$ Mean $= \frac{2500}{50} = 50$ Observations: The mean of the first n odd numbers is always n . For 50 -odd numbers, the mean is 50. (iii) Mean of the first 50 multiples of 4 .

The first 50 multiples of $4 = 4, 8, 12$ , 16, ......, 200 Sum of multiples of 4 $= 4 + 8 + 12 + \dots + 200$ $= 4 (1 + 2 + 3 + \dots + 50)$ $= 4 \times (1275) = 5100$ Mean $= \frac{5100}{50} = 102$ Observation: The mean of the first n multiples of 4 is 4 times the mean of the first n natural numbers. Mean $= 4 \times 25.5 = 102$

2. The dot plot below shows a collection of data and its average; but one dot is missing. Mark the missing value so that the mean is 9 (as shown below.)

Sol. From the dot plot, the data values4, 7, 8, $8, 9, 9, 9, 9, 9, 11$ , xNumber of observations = 11Mean = 9Sum of data values $= 4 + 7 + 8 + 8 + 9 + 9 + 9 + 9 + 9 + 11 = 83$ Total sum $= 9 \times 11 = 99$ Missing number $= 99 - 83 = 16$

3. Sudhakar, the class teacher, asks Shreyas to measure the heights of all 24 students in his class and calculate the average height. Shreyas informs the teacher that the average height is 150.2 cm . Sudhakar discovers that the students were wearing uniform shoes when the measurements were taken and the shoes add 1 cm to the height. (i) Should the teacher get all the heights measured again without the shoes to find the correct average height? Or is there a simpler way? (ii) What is the correct average height of the class? (a) 174.2 cm (b) 126.2 cm (c) 150.2 cm (d) 149.2 cm (e) 151.2 cm (f) None of the above (g) Insufficient information

Sol. (i) The teacher does not need to remeasure all the heights. Since the shoes add 1 cm to every student's height, the average height with shoes is 1 cm more than the actual average height. So, the correct average height is 1 cm less than the measured average height. (ii) Correct average height = $150.2 cm - 1 cm = 149.2 cm$ $∴$ Option (d) is correct

4. The three dot plots below show the length, in minutes, of songs of different albums. Which of these has a mean of5.57 minutes? Explain how you arrived at the answer.

B

C

Sol. We can determine it by examining the dot plots. In $dot$ plot A , most song lengths are between 5 and 6.5 minutes, so the mean is likely around 5.57 minutes. In dot plots B and C , all song lengths are below 5.57 minutes, so their means cannot exceed 5.57. Therefore, $dot plot A$ is the one with a mean of 5.57 minutes.

Check:

For Album A

Data values $= 5, 5, 5.25, 5.5, 5.75, 6, 6.5$ Mean $= \frac{5 + 5 + 5.25 + 5.5 + 5.75 + 6 + 6.5}{7}$ $= \frac{39}{7} = 5.57$ minutes

For Album B

Data values $= 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5$ Mean $= \frac{0.5 + 0.75 + 1.5 + 1.5 + 2 + 3.75 + 4.25 + 5}{8}$ $= \frac{19.25}{8} = 2.41$ minutes

For album C

Data values $= 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5$ Mean $= \frac{3.5 + 3.5 + 3.5 + 4 + 4 + 4 + 4.25 + 4.5}{8}$ $= \frac{31.25}{8} = 3.9$ minutes Album (A) has a mean of 5.57 minutes

5. Find the median of $8, 10, 19, 23, 26, 34$ , 40, 41, 41, 48, 51, 55, 70, 84, 91, 92. (i) If we include one value to the data (in the given list) without affecting the median, what could that value be? (ii) If we include two values to the data without affecting the median what could the two values be? (iii) If we remove one value from the data without affecting the median what could the value be?

Sol. The given data set is $8, 10, 19, 23, 26, 34$ , 40, 41, 41, 48, 51, 55, 70, 84, 91, 92

The number of observations ( n ) $= 16$ (even) Median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ term $=$ average of $8^{th}$ and $9^{th}$ terms $= \frac{41 + 41}{2} = \frac{82}{2} = 41$ (i) To keep the median 41 with an odd number of values ( $n = 17$ ), the new value must be placed at the median itself in the order list for making the middle value the 9th value. So, the value could be 41. (ii) With $n = 18$ (even) the median is the average of the 9th and 10th values. To keep the median 41 , we need to add two numbers whose sum is 82 . For this, one value should be less than 41 and the other greater than 41.For example, the two values could be 40 and 42 . (iii) With $n = 15$ (odd) the median is the $8^{th}$ value that is 41 . So, we can remove another 41 from the ordered list.

6. Examine the statements below and justify if the statement is always true, sometimes true, or never true. (i) Removing a value less than the median will decrease the median. (ii) Including a value less than the mean will decrease the mean. (iii) Including any 4 values will not affect the median. (iv) Including 4 values less than the median will increase the median.

Sol. (i) Sometimes true: Removing a value less than the median can decrease the median. But if the data set has an even number of elements and the removed value is below the lower of the two middle values, the median may stay the same. (ii) Always true: The mean is the sum of all values divided by the number of values. Adding a value less than the mean decreases the total sum less than proportionally to the increase in the number of values, thus decreasing the mean. (iii) Sometimes true: Including any 4 values could shift the median depending on whether those values are above or below the median and on the original number of observations. (iv) Never true: Including values less than the median will either keep the median the same or decrease it, but it will never increase it.

7. The mean of the numbers $8, 13, 10, 4, 5$ , $20, y, 10$ is 10.375 . Find the value of $y$ .

Sol. Numbers $= 8, 13, 10, 4, 5, 20, y, 10$ Total numbers $= 8$ Mean $= 10.375$ Total Sum $= 10.375 \times 8 = 83$ Sum of numbers $= 8 + 13 + 10 + 4 + 5 + 20 + y + 10$ $= 70 + y$ So, $83 = 70 + y$ $y = 83 - 70 = 13$

8. The mean of a set of data with 15 value is 134. Find the sum of the data.

Sol. Mean $= 134$ Total values $= 15$ Sum of data $= 134 \times 15 = 2010$

9. Consider the data: $12, 47, 8, 73, 18, 35$ , 39, 8, 29, 25, p. Which of the following number(s) could be p if the median of this data is 29 ? (i) 10
(ii) 25 (iii) 40 (iv) 100 (v) 29 (vi) 47 (vii) 30

Sol. Arranging the data (excluding p) $8, 8, 12, 18, 25, 29, 35, 39, 47, 73$ Total observations = 11 (odd) Median $= (\frac{n + 1}{2})^{th}$ term $= 6^{th}$ term $= 29$ So, p must be $\geq 29$ to keep the value 29 in the middle.

So possible values of p are (iii) 40, (iv) 100, (v) 29, (vi) 47, and (vii) 30.

10. The number of times students rode their cycles in a week is shown in the dot plot below. Four students rode their cycles twice in that week.

(i) Find the average number of times students rode their cycles. (ii) Find the median number of times students rode their cycles. (iii) Which of the following statements are valid? Why? (a) Everyone used their cycle at least once. (b) Almost everyone used their cycle a few times. (c) There are some students who cycled more than once on some days. (d) Exactly 5 students have used their cycles more than once on some days. (e) The following week, if all of them cycled 1 more time than they did the previous week, what would be the average and median of the next week's data?

Sol. From the dot plot Data

Number of times students rode their cycles (x)	Number of students(f)
0	3
1	1
2	4
3	7
4	7
5	5
6	4
7	6
8	3
9	-
10	2

(i) Sum $= (0 \times 3) + (1 \times 1) + (2 \times 4) + (3 \times 7) + (4 \times 7) + (5 \times 5) + (6 \times 4) + (7 \times 6) + (8 \times 3) + (10 \times 2) = 0 + 1 + 8 + 21 + 28 + 25 + 24 + 42 + 24 + 20 = 193$

Total numbers of students $= 42$ Average $= \frac{193}{42} = 4.59$ times (ii) Since total number of students $= 42$ Since number of observations is even, we use the average of the two middle numbers

Now, Median = $\frac{( \frac{n}{2} ) ^{th} observation + ( \frac{n + 1}{2} ) ^{th} observation}{2}$ $= \frac{( \frac{42}{2} ) ^{th} observation + ( \frac{42}{2} + 1 ) ^{th} observation}{2}$

$= \frac{2 1 ^{st} observation + ( 21 + 1 ) ^{th} observation}{2} = \frac{2 1 ^{st} observation + 2 2 ^{nd} observation}{2}$

Thus, median is the average of the $21^{st}$ and $2 2^{nd}$ values

Counting the dots from left to right, both the $2 1^{st}$ and $2 2^{nd}$ dots fall in the "4" column.

The median is 4. (iii)(a)Invalid: Three students rode 0 times. (b)Valid: The heavy clustering in the middle shows almost everyone rode a few times. (c) Valid: Three are only 7 days in a week. Some students recorded 8 and 10 rides, meaning they had to ride multiple times in a single day to reach that number. (d)Not valid: More than 5 students have used their cycles more than once on some days. (e) If everyone cycles 1 more time next week, the entire data set simply shifts up by 1 .

Thus, New average $=$ old average +1 $= 4.59 + 1 = 5.59$ (approx.) and New median $=$ old median $= 4 + 1 = 5$

11. A dart-throwing competition was organised in a school. The number of throws participants took to hit the bull's eye (the centre circle) is given in the table below. Describe the data using its minimum, maximum, mean and median.

No. of Trials	No. of Students
1	1
2	0
3	0
4	1
5	4
6	9
7	12
8	15
9	10
10	10

Sol. Minimum: the lowest number of throws to hit the bullseye was 1 .

Maximum: the height number of throws was 10 .

Here, Total number of students

$= 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62$ and Number of trails $= 1 \times 1 + 2 \times 0 + 3 \times 0 + 4 \times 1 + 5 \times 4 + 6 \times 9 + 7 \times 12 + 8 \times 15 + 9 \times 10 + 10 \times 10 = 473$

Finding mean

Mean $= \frac{Total number of trials}{Total number of students}$ $= \frac{473}{62} = 7.62$ (approx.)

Since total number of students $= 62$ Since number of observations is even, We use the average of two middle numbers

Now, Median = $\frac{( \frac{n}{2} ) ^{th} observation + ( \frac{n + 1}{2} ) ^{th} observation}{2}$ $= \frac{( \frac{62}{2} ) ^{th} observation + ( \frac{62}{2} + 1 ) ^{th} observation}{2}$ $= \frac{3 1 ^{st} observation + ( 31 + 1 ) ^{th} observation}{2}$ $= \frac{3 1 ^{st} observation + 3 2 ^{nd} observation}{2}$ Thus, median is the average of the $3 1^{st}$ and $3 2^{nd}$ values.

If we keep a running total of the frequencies from left to right, the $3 1^{st}$ and $3 2^{nd}$ values both fall inside the column for 8 throws.

Thus, median is 8

No. of trials	No. of students	Successive addition
1	1	1
2	0	$1 + 0 = 1$
3	0	$1 + 0 = 1$
4	1	$1 + 1 = 2$
5	4	$2 + 4 = 6$
6	9	$6 + 9 = 15$
7	12	$15 + 12 = 27$
8	15	$27 + 15 = 42$
9	10	$42 + 10 = 52$
10	10	52 + 10 = 62

3.0Key Concepts of Chapter 5 Tales by Dots and Lines Exercise 5.1

Finding the mean of a data set
Understanding mean as a balancing point
Finding missing values using a given mean
Finding the median of an ordered data set
Studying changes in mean and median when data values change
Interpreting dot plots using mean and median

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 5 Exercise 5.1

Calculates the mean of a given data set.
Identifies the mean as the balancing point of a data set.
Determines missing values when the mean is known.
Finds the median of an ordered data set.
Examines changes in the mean and median when data values are added or removed.
Interprets dot plots using mean and median.
Compares the effect of individual data values on the mean and median.

NCERT Class 8 Maths Ch. 5 Tales by Dots and Lines Other Exercises:-

Exercise 5.1

Exercise 5.2